A-level chemistry

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ok im lost again.
isnt negative... if its going in the opposite direction.. orrr if thers an increase in temp?
and delta H energy change/mole?
If a system releases heat(exothermic reaction) the energy of the system logically decreases so the enthalpy is negative.
If a system absorbs energy, it's energy level increases, hence the positive enthalpy.
Delta H - change in enthalpy
Delta T - change in temperature
Delta n -change in the number of moles...
That's how I was taught..:skeleton:
 
Let's see if we're talking about the same thing...

Here's a simple chemical reaction, the burning of methane.

CH4(g)+2O2(g) --> CO2(g) + 2H2O(l, just to make it easier)

the enthalpy of a recation is defined as
the enthalpy of the products minus the enthalpy of the reactants(reagents)

1 mol × (-393 kJ/mol) + 2 mol × (-286 kJ/mol)
minus
1 mol × ( -74,8 kJ/mol) + 2 mol × 0 (because the enthalpy of an elemnet (oxygen) equals zero)
equals
-890 kJ

890kJ is the amount of energy released from burning 1 mole of methane. As you know burning produces heat - releases energy.

Are we talking about the same thing then?:muddlehea
 
woahhh :S
uv reli lost me. sorry, i must sound reli unintelligent but i dont get it.

i reli appreciate your responses.

isnt it:
Q= change in temp * mass * specific heat capacity ~4.18

thenn
Delta H = Q/moles

and if its exothermic, delta H is negative?
 
woahhh :S
uv reli lost me. sorry, i must sound reli unintelligent but i dont get it.

i reli appreciate your responses.

isnt it:
Q= change in temp * mass * specific heat capacity ~4.18

thenn
Delta H = Q/moles

and if its exothermic, delta H is negative?
That sounds more like physics to me.
Yes, Delta Q (change in energy) = Delta T (chnage in temperature) × c (specific heat capacity) × m (mass).
letters my be different in the UK, I think they're different (some of them) in the US.
i'm not sure what 4.18 signifies...
I've been taught that Delta H signifies a change in enthalpy, wikipedia says the same.
Delta H is not Q/moles. Standard enthalpy(enthalpy per one mole) is Q/moles. You get an enthalpy(or change in it) by multiplying the standard one with the number of moles, as shown in the reaction above.
Yes, enthalpy of an exothermic reaction is negative.

sorry if I'm confusing you.:uuh:
 
Last edited:
I figured out what al_islam has been saying
ps remember to use a negative sign for the reaction going in the oppostie direction.
When a reaction goes in the opposite direction all enthalpies change +/- signs.
It's logical, when you think about it, samia. If a reaction going in one direction released energy, the reversal of the reaction will have to absorb it, exactly the same amount of it. And vise versa. therefore, the enthalpy changes its sign (+,-).
 
Now, if you have to claculate an enthalpy for a reaction you just have to calculate the entalpies of every compund and sum it up as in the reaction above.
If the reaction goes in the opposite direction and the data you have (enthalpies fo every compound) refer to the original (left to right) direction, you need to change signs (+/-). As easy as that.

I'm still not sure what the triangles are though...:playing:
 
I'm still not sure what the triangles are though...:playing:

The triangle is just a way of expressing the equations you have written in another form.

this makes it easier to see the equation and the -/+ values, and which way is negative and which is positive.

CH4(g)+2O2(g) --> CO2(g) + 2H2O(l, just to make it easier)

the enthalpy of a recation is defined as
the enthalpy of the products minus the enthalpy of the reactants(reagents)

1 mol × (-393 kJ/mol) + 2 mol × (-286 kJ/mol)
minus
1 mol × ( -74,8 kJ/mol) + 2 mol × 0 (because the enthalpy of an elemnet (oxygen) equals zero)
equals
-890 kJ

890kJ is the amount of energy released from burning 1 mole of methane. As you know burning produces heat - releases energy.


CH4 + 2O2





CO2 2H2O




Imagine arrows to the CO2 and 2H2O, I cant seem to get them here.

The enthalpy change that you need to work out can be clearly seen.

If you need to go in the oppostie direction, then you must use the OPPOSITE temperature sign. So if its positive then you use negative...and the other way around.

Also remember to use the equation.

Delta H is a change in enthalpy - energy. Delta T is a change in temperature
.

OOPs sorry, i got it the other way around....mustve been tired that day, thanks for the correction 'whatsthe point'.

if u dont get it, PM me the question...ill hav a crack at it for you....i was jus tellin the theory behind it.
 
I'm still not sure what the triangles are though...:playing:

The triangle is just a way of expressing the equations you have written in another form.

this makes it easier to see the equation and the -/+ values, and which way is negative and which is positive.

CH4(g)+2O2(g) --> CO2(g) + 2H2O(l, just to make it easier)

the enthalpy of a recation is defined as
the enthalpy of the products minus the enthalpy of the reactants(reagents)

1 mol × (-393 kJ/mol) + 2 mol × (-286 kJ/mol)
minus
1 mol × ( -74,8 kJ/mol) + 2 mol × 0 (because the enthalpy of an elemnet (oxygen) equals zero)
equals
-890 kJ

890kJ is the amount of energy released from burning 1 mole of methane. As you know burning produces heat - releases energy.


CH4 + 2O2





CO2 2H2O




Imagine arrows to the (CO2) and (2H2O) coming from (CH4 + 202) - 2 separate ones: 1 to CO2 and one to 2H2O.

AND ALSO from (CO2) to (2H2O) - I cant seem to get them here.

The enthalpy change that you need to work out can be clearly seen.

Insert the temperature values given for each reaction, and work out the one you need to find.

If you need to go in the oppostie direction, then you must use the OPPOSITE temperature sign. So if its positive then you use negative...and the other way around.

Also remember to use the equation.

Delta H is a change in enthalpy - energy. Delta T is a change in temperature
.

OOPs sorry, i got it the other way around....mustve been tired that day, thanks for the correction 'whatsthe point'.

if u dont get it, PM me the question...ill hav a crack at it for you....i was jus tellin the theory behind it.
 
Last edited:
i think

i finally get it

thank u so much for ur help!

Its ok.

Just show me the question and ur answer.

Ill check it for you.

But its not so bad.

Any more questions just ask.....Nuffield needs work....good to hear ur revising.

Good luck.
 
Its ok.

Just show me the question and ur answer.

Ill check it for you.

But its not so bad.

Any more questions just ask.....Nuffield needs work....good to hear ur revising.

Good luck.

i read that just now ^^

its driving me round the bend.
thanks for the offer :)
 
:sl:
I did AS Salters OCR Chemistry, I hated every lesson of it, I'm so glad I dropped it at the end. Now Computing all the way for me :D
 

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